How to evaluate $\iiint_{\mathbb{R}^3} \exp \left(-x^2 -y^2-z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z$

Question

How to evaluate $$ \iiint_{\mathbb{R}^3} \exp \left( -x^2 -y^2 -z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z, $$

My Attempt:

If we use the transformation to spherical polar coordinates $R, \theta, \phi$, then we have $$ \begin{align} x &= R \sin \theta \cos \phi, \\ y &= R \sin \theta \sin \phi, \\ z &= R \cos \theta, \end{align} $$ where $0 \leq r < +\infty$, $0 \leq \theta \leq \pi$, and $0 \leq \phi < 2 \pi$.

And thus the Jacobian determinant

\begin{align} &\frac{ \partial (x, y, z) }{ \partial (R, \theta, \phi ) } = \left\lvert \begin{matrix} \frac{\partial x}{\partial R } & \frac{\partial x }{ \partial \theta } & \frac{ \partial x }{ \partial \phi} \\ \frac{\partial y}{\partial R } & \frac{\partial y }{ \partial \theta } & \frac{ \partial y }{ \partial \phi} \\ \frac{\partial z }{\partial R } & \frac{ \partial z }{ \partial \theta } & \frac{ \partial z }{ \partial \phi} \end{matrix} \right\rvert \\ &= \left\lvert \begin{matrix} \sin \theta \cos \phi & R \cos \theta \cos \phi & - R \sin \theta \sin \phi \\ \sin \theta \sin \phi & R \cos \theta \sin \phi & R \sin \theta \cos \phi \\ \cos \theta & -R \sin \theta & 0 \end{matrix} \right\rvert \\ &= R^2 \sin \theta \left\lvert \begin{matrix} \sin \theta \cos \phi & \cos \theta \cos \phi & - \sin \phi \\ \sin \theta \sin \phi & \cos \theta \sin \phi & \cos \phi \\ \cos \theta & - \sin \theta & 0 \end{matrix} \right\rvert \\ &= R^2 \sin \theta \left( -\sin \phi \left\lvert \begin{matrix} \sin \theta \sin \phi & \cos \theta \sin \phi \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert - \cos \phi \left\lvert \begin{matrix} \sin \theta \cos \phi & \cos \theta \cos \phi \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert \right) \\ &= R^2 \sin \theta \left( -\sin^2 \phi \left\lvert \begin{matrix} \sin \theta & \cos \theta \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert - \cos^2 \phi \left\lvert \begin{matrix} \sin \theta & \cos \theta \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert \right) \\ &= R^2 \sin \theta \left( -\sin^2 \phi (-1) - \cos^2 \phi (-1) \right) \\ &= R^2 \sin \theta. \end{align}

And, therefore we have

\begin{align} \iiint_V &\exp \left( -x^2 -y^2 -z^2 \right) \, dx\, dy\, dz = \int_0^{2\pi} \int_0^\pi \int_0^\infty \exp\left( -R^2 \right)R^2 \sin \theta \, dR \, d\theta \, d\phi \\ &= 4 \pi \int_0^\infty R^2 \exp \left( - R^2 \right) \, dR \\ &= 4 \pi \left( - \frac{ R \exp \left( -R^2 \right) }{2 } \right)_{R=0}^{R=\infty} + 2 \pi \int_0^\infty \exp \left( -R^2 \right) \, dR \\ &= 2 \pi \int_0^\infty \exp \left( -R^2 \right) \, dR \\ &= \end{align}

Is what I have done so far correct and clear enough? If so, then how to proceed from here? How to evaluate $$ \int_0^\infty \exp \left( -R^2 \right) \, dR ? $$

Answer 1

The last integral is Gaussian integral

$$\int_0^{+\infty}e^{-x^2}dx = \dfrac{\sqrt{\pi}}{2}$$

There are many ways to evaluate this integral, and here is one way https://en.wikipedia.org/wiki/Gaussian_integral

Answer 2

The integral $\int_{0}^{\infty}e^{-R^2}dR$ is the Gaussian Integral.

It is very well known, the wikipedia link I have given above also contains a derivation. It equals-

$\int_{0}^{\infty}e^{-R^2}dR=\frac{\sqrt{\pi}}{2}$

So for your original integral,

$\begin{align}&\int\int\int_{\mathbb{R}^3}e^{-(x^2+y^2+z^2)}dxdydz\\&=2\pi\int_{0}^{\infty}e^{-R^2}dR\\&=2\pi\frac{\sqrt{\pi}}{2}\\&=\pi\sqrt{\pi}\\&=\pi^{\frac{3}{2}}\end{align}$